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Finding one root $x_1$ of the quintic equation $x^5 + x = -a$ by using the Bring radical is described on Wikipedia.

The root is $x_1 = -a +a^5 -5a^9+35a^{13}+ \ldots$ , and it is found by reversion of the Taylor series for $f(x) = x^5 + x$.

How do we find the other roots of this quintic in series representation?

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2 Answers 2

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The answer is contained in the paper

The Bring-Jerrard normal form is a particular example of a trinomial, and Eagle gives the general solution.

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You can get all the roots for any trinomial by using the reversion theorem depending on the radius of convergence. Surpsingly, so such solution exits online for the general trinomial.

so for example $x^5-2x^2+z=0$ we have two roots with a Euclidian norm of less than one and three roots greater than one.

we would have two partitions:

$f(x)=x^5-2x^2$ which gives three roots

you you would apply it around the region of $2^{1/3}$ which gives a real root and two complex ones. For small z, one can check that these are close to the actual answer

or

$f(x)=-2x^2+z$ and gives the remaining two

This bypasses the resolvent method, which results in multiple hypergeometric series for each root. You only need one hypergeometric function for each root.

this combines this

https://arxiv.org/pdf/0910.2957.pdf

with

https://en.wikipedia.org/wiki/Bring_radical

the first paper gives the formula for the small roots (14,15)

the wiki page (Glasser's derivation) will give the others

It's very hard to find good summaries about how to apply the Lagrange inversion to actually derive these. If anyone can prove these formulas it would be great. You have to make some transformations so $x^5-2x^2+z=0$ becomes $1+y^2+y^5(1/z)(-z/2)^{5/2}$ and the root of the original equation is $y(-z/2)^{1/2}$. It can be done.

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